In particular, I'm trying to find the cyclic subgroups of $D_8=\{1,r,r^2,r^3,s,sr,sr^2,sr^3\}$, the dihedral group. Seems like quite a hassle to check every possible subset of $D_8$ as a generator. I've gone through each one element generator to see what cyclic subgroups I get from that.

Is there any time-saving insight that says it's good enough to check the one element generators (in general, or just for this problem)?


  • $\begingroup$ What specifically do you mean by 'one-element generators'? It's true that every cyclic group is generated by a single element, so you only have to check the individual elements to see what groups they generate - but in the case you give, for instance, that doesn't mean that you only have to test $r$ and $s$ as generators... $\endgroup$ – Steven Stadnicki Feb 15 '17 at 23:33
  • $\begingroup$ I meant I checked each element of $D_8$. Ok, I missed that in reading the definition of cyclic groups. Thanks a bunch! $\endgroup$ – manofbear Feb 15 '17 at 23:34
  • $\begingroup$ One way to save further work is you know the only candidates besides the trivial group are $Z_2$ and $Z_4$ since the order of the group is $8$ and Lagrange's thm implies the order of the subgroups must be $1,2,4,$ or $8.$ $Z_8$ is out cause $D_8$ has $8$ elements and is not $Z_8.$ Then you just find one element that generates $Z_2$ and one that generates $Z_4$ and don't have to bother checking any of the others. $\endgroup$ – spaceisdarkgreen Feb 15 '17 at 23:39
  • $\begingroup$ @space While you're correct that those are the only possible isomorphism classes of cyclic subgroups (except for the trivial group), there are $5$ copies of $\mathbb Z_2$ and I believe the goal is to find each specific one. In that case you still need to check every element. $\endgroup$ – Matt Samuel Feb 15 '17 at 23:42
  • $\begingroup$ @MattSamuel true, I think that's probably the correct interpretation of the question. $\endgroup$ – spaceisdarkgreen Feb 15 '17 at 23:47

A cyclic group is by definition a group with one generator. An algorithm for solving this problem would be to take each element and exponentiate it until you get the identity. The sequence of elements you encountered along the way forms a cyclic subgroup.

While it may seem like an arduous task to check every element, I suspect that once you get down to it you will discover that there are ways to save work.

  • $\begingroup$ Excellent, I see it in the definition now. Thanks for the prompt response $\endgroup$ – manofbear Feb 15 '17 at 23:35
  • $\begingroup$ @manofbear No problem. $\endgroup$ – Matt Samuel Feb 15 '17 at 23:36

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